Solve the following:

There are 6 positive and 8 negative numbers. Four numbers are chosen at random, without replacement, and multiplied. Find the probability that the product is a positive number.

#### Solution

Let event A: Four positive numbers are chosen.

event B: Four negative numbers are chosen

event C: Two positive and two negative numbers are chosen.

Since four numbers are chosen without replacement,

n(A) = 6 × 5 × 4 × 3 = 360

n(B) = 8 × 7 × 6 × 5 = 1680

In event C, four numbers are to be chosen without replacement such that two numbers are positive and two numbers are negative. This can be done in following ways:

+ + – – OR + – + – OR + – – + OR – + – + OR – – + + OR – + + –

∴ n(C) = 6 × 5 × 8 × 7 + 6 × 8 × 5 × 7 + 6 × 8 × 7 × 5 + 8 × 6 × 7 × 5 + 6 × 5 × 8 × 7 + 8 × 6 × 5 × 7

= 6 × (8 × 7 × 6 × 5)

= 10080

Here, total number of numbers = 14

∴ n(S) = 14 × 13 × 12 × 11 = 24024

Since A, B, C are mutually exclusive events,

Required probability = P(A) + P(B) + P(C)

= `("n"("A"))/("n"("S")) + ("n"("B"))/("n"("S")) + ("n"("C"))/("n"("S"))`

=`(360 + 1680 + 10080)/24024`

= `505/1001`